`.^(23)` Na is the more stable isotope of Na. Find out the process by which `._(11)^(24)` Na can undergo radioactive decay.

Options

A. `beta^(-)` emssion
B. `alpha`- emission
C. `beta^(+)` emssion
D. K electron capture

Answer

A

Explanation

n/p ratio of `.^(24)Na` nuclide is 13/11 i.e., greater than unity and hence radioactive. To achieve stability, it would tend to adjust its n/p ratio to the proper value of unity. This can be done by breaking a neutron into proton and electron.
`._(0)n^(1) rarr ._(+1)P^(1) + ._(-1)e^(0) or beta^(-)`
The proton will stay inside the nucleus whereas electron which cannot exist in the nucleus, will be emitted out as `beta-` ray